According to the Rational Zeros Theorem, a polynomial with integer coefficients has a rational root in the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
We have the polynomial:
[tex]x^3+4x^2-74x-77[/tex]a.
The factors of the constant term are:
[tex]\pm1,\operatorname{\pm}7,\operatorname{\pm}11,\operatorname{\pm}77[/tex]And the factors of the leading coefficient are -1 and +1.
Thus, the possible zeros are (comma-separated):
-1, +1, -7, +7, -11, +11, -77, +77
b. Now we use synthetic division to find the rational root from the list above.
Let's start with x = +1:
1 4 -74 -77
+ 1 +1 5 -69
--------------------------------
1 5 -69 -146
Since the last result is not zero, x = 1 is not a root of the polynomial.
Let's try x = -1:
1 4 -74 -77
- 1 -1 - 3 +77
--------------------------------
1 3 -77 0
The last result is 0 thus, x = -1 is a rational zero of the function.
One rational zero of the given function is -1
c.
The zero we have found in part b helps us to find the other two zeros. The three coefficients remaining in the last row are the coefficients of the quadratic polynomial a = 1, b = 3, c = -77 which can be solved by using the formula:
[tex]$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$[/tex]Substituting:
[tex]\begin{gathered} x=\frac{-3\pm\sqrt{3^2-4*1*\left(-77)\right?}}{2*1} \\ x=\frac{-3\pm\sqrt{9+308}}{2}=\frac{-3\pm\sqrt{317}}{2} \end{gathered}[/tex]We have two different zeros:
[tex]\begin{gathered} x=\frac{-3+\sqrt{317}}{2} \\ x=\frac{-3-\sqrt{317}}{2} \end{gathered}[/tex]